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Let the numbers be x and y

Step 1:

The difference of numbers is 12

=> x - y = 12 .........................................(1)

and

Three times one of these numbers exceeds two times the other number by 10.

=> 3 times one number - 2 times other = 10

=> 3x - 2y = 10 ..........................................(2)

Step 2:

(1) => x = 12 + y .......................................... (3)

Put x = 12 + y in equation (2)

=> 3(12 + y) - 2y = 10

=> 36 + 3y - 2y = 10

=> y = 10 - 36 = - 26

Step 3:

Put y = - 26 in equation (3)

=> x = 12 + (- 26)

=> x = 12 - 26 = -14

**Answer**: x = - 14 and y = - 26

Step 1:

The difference of numbers is 12

=> x - y = 12 .........................................(1)

and

Three times one of these numbers exceeds two times the other number by 10.

=> 3 times one number - 2 times other = 10

=> 3x - 2y = 10 ..........................................(2)

Step 2:

(1) => x = 12 + y .......................................... (3)

Put x = 12 + y in equation (2)

=> 3(12 + y) - 2y = 10

=> 36 + 3y - 2y = 10

=> y = 10 - 36 = - 26

Step 3:

Put y = - 26 in equation (3)

=> x = 12 + (- 26)

=> x = 12 - 26 = -14

Given, 32x - 46 = 12x + 20

Solve for x,

32x - 46 = 12x + 20

Subtract 12x from both sides

=> 32x - 46 - 12x = 12x + 20 - 12x

=> 20x - 46 = 20

Add 46 to each side of the equation

=> 20x - 46 + 46 = 20 + 46

=> 20x = 66

Divide each side by 20

=> x = $\frac{66}{20}$

or x = $\frac{33}{10}$

Solve for x,

32x - 46 = 12x + 20

Subtract 12x from both sides

=> 32x - 46 - 12x = 12x + 20 - 12x

=> 20x - 46 = 20

Add 46 to each side of the equation

=> 20x - 46 + 46 = 20 + 46

=> 20x = 66

Divide each side by 20

=> x = $\frac{66}{20}$

or x = $\frac{33}{10}$

Given quadratic equation, x^{2} - 13x + 42

Factorized the equation

x^{2} - 13x + 42 = x^{2} - 7x - 6x + 42

= x(x - 7) - 6(x - 7)

= (x - 6)(x - 7)

=> x^{2} - 13 + 42 = (x - 6)(x - 7) ^{}

Factorized the equation

x

= x(x - 7) - 6(x - 7)

= (x - 6)(x - 7)

=> x

$\frac{2^8}{2^5}$ x $\frac{3^4}{3^7}$ = 2^{8 - 5} x 3^{4 - 7}

[Using law of indices, $\frac{x^m}{x^n}$ = x^{m - n }]

=> $\frac{2^8}{2^5}$ x $\frac{3^4}{3^7}$ = 2^{3} x 3^{- 3
}= $\frac{2^3}{3^3}$

= $\frac{8}{27}$

=> $\frac{2^8}{2^5}$ x $\frac{3^4}{3^7}$ = $\frac{8}{27}$

[Using law of indices, $\frac{x^m}{x^n}$ = x

=> $\frac{2^8}{2^5}$ x $\frac{3^4}{3^7}$ = 2

= $\frac{8}{27}$

=> $\frac{2^8}{2^5}$ x $\frac{3^4}{3^7}$ = $\frac{8}{27}$